Simultaneous Equations Quiz
Test your knowledge of Simultaneous Equations for GCSE Maths, with this quiz.
This quiz consists of 15 questions, including multiple-choice and short-answer questions on the topic of Simultaneous Equations for GCSE Maths.
For multiple-choice questions, choose the correct answer. Scroll down to begin the quiz.
Questions
What is the solution to the system: x + y = 10, x − y = 2?
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What method is best suited when one variable can be easily eliminated by adding or subtracting the equations?
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What is the solution to: 2x = 6 and x + y = 7?
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If you substitute a value of x into one equation to findy, which method is this?
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What is the solution to the system: x + y = 7, x = 2y?
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Solve: x + y = 8 and x − y =4
Answer:
Add equations: $2x = 12 \Rightarrow x = 6$
Then: $6 + y = 8 \Rightarrow y = 2$
Solution: $x = 6, y = 2$
Solve using substitution: y = 2x + 1 and x + y = 10
Answer:
Substitute: $x + (2x + 1) = 10 \Rightarrow 3x + 1 = 10 \Rightarrow x = 3$
Then: $y = 2(3) + 1 = 7$
Solution: $x = 3, y = 7$
Solve: 3x + 2y = 12 and 2x − y = 1
Answer:
Multiply second equation by 2:
$4x - 2y = 2$
Add to first: $3x + 2y + 4x - 2y = 12 + 2 \Rightarrow 7x = 14 \Rightarrow x = 2$
Then: $2(2) - y = 1 \Rightarrow 4 - y = 1 \Rightarrow y = 3$
Solution: $x = 2, y = 3$
Solve: 2x + y = 5 and x − y =1
Answer:
Add equations: $3x = 6 \Rightarrow x = 2$
Then: $2 + y = 5 \Rightarrow y = 3$
Solution: $x = 2, y = 3$
Solve: x + y = 6 and x − y =2
Answer:
Add: $2x = 8 \Rightarrow x = 4$
Then: $4 + y = 6 \Rightarrow y = 2$
Solution: $x = 4, y = 2$
Solve: 2x + 3y =12 and x − y = 1
Answer:
From second: $x = y + 1$
Sub into first: $2(y + 1) + 3y = 12 \Rightarrow 2y + 2 + 3y = 12 \Rightarrow 5y = 10 \Rightarrow y = 2$
Then: $x = 2 + 1 = 3$
Solution: $x = 3, y = 2$
Solve graphically: y= 2x + 1 and y=−x+4
Answer:
Set equal: $2x + 1 = -x + 4 \Rightarrow 3x = 3 \Rightarrow x = 1$, then $y = 3$
Solution: $x = 1, y = 3$
Solve: x = 3y and x + y = 16
Answer:
Sub: $3y + y = 16 \Rightarrow 4y = 16 \Rightarrow y = 4$, then $x = 12$
Solution: $x = 12, y = 4$
Solve: x + 3y = 11 and 2x − y = 1
Answer:
Multiply second by 3: $6x - 3y = 3$
Add to first: $x + 3y + 6x - 3y = 11 + 3 \Rightarrow 7x = 14 \Rightarrow x = 2$
Then: $2 + 3y = 11 \Rightarrow y = 3$
Solution: $x = 2, y = 3$
Solve: x − 2y = −1 and 3x + y = 9
Answer:
Multiply first by 1: $x = 2y - 1$
Sub: $3(2y - 1) + y = 9 \Rightarrow 6y - 3 + y = 9 \Rightarrow 7y = 12 \Rightarrow y = \frac{12}{7}$
Then: $x = 2(\frac{12}{7}) - 1 = \frac{17}{7}$
Solution: $x = \frac{17}{7}, y = \frac{12}{7}$
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